Multiplication in a -algebra is continuous, and so is the map in . To show that is continuous, it therefore suffices to show that the map is continuous. This map is composed of the map and the map for .
The former of these maps is continuous because the involution and multiplication is continuous. It now suffices to show that is continuous on any bounded set. But this follows from Lemma 1.2.5. because each bounded subset of is contained in , where and
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Aside:
is the set of self-adjoints in with spectrum contained in .
Lemma 1.2.5. Says that any continuous function on , then is continuous.
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If is a unitary element in , then and so .
Let be given, and put for .This is the homotopy between z and its normalization
Then and . Since is positive and invertible, there exists for which .
For each , we have
Thus and are invertible. The map is continuous, and so in